McqMate
These multiple-choice questions (MCQs) are designed to enhance your knowledge and understanding in the following areas: Electrical Engineering .
| 101. |
The frequency of oscillation in Gunn diode is given by: |
| A. | vdom/ leff |
| B. | leff/ vdom |
| C. | leff/ wvdom |
| D. | none of the mentioned |
| Answer» A. vdom/ leff | |
| Explanation: in gunn oscillation mode, the frequency of oscillation is given by vdom/ leff, where vdom is the domain velocity, leff is effective length that the domain moves from the time it is formed until the time a new domain is formed. | |
| 102. |
In Gunn diode oscillator, the Gunn diode is inserted into a waveguide cavity formed by a short circuit termination at one end |
| A. | true |
| B. | false |
| Answer» A. true | |
| Explanation: the gunn diode is mounted at the centre of the broad wall of a shorted waveguide since for the dominant te10 mode; the electric field is maximum at the centre. | |
| 103. |
In a Gunn diode oscillator, the electron drift velocity was found to be 107 cm/second and the effective length is 20 microns, then the intrinsic frequency is: |
| A. | 5 ghz |
| B. | 6 ghz |
| C. | 4 ghz |
| D. | 2 ghz |
| Answer» A. 5 ghz | |
| Explanation: the intrinsic frequency for a gunn oscillator is given by vd/l. here vd is the drift velocity and l is the effective length. substituting the given values in the above equation, intrinsic frequency is 5 ghz. | |
| 104. |
The material used to fabricate IMPATT diodes is GaAs since they have the highest efficiency in all aspects. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: impatt diodes can be fabricated using silicon, germanium, gaas or indium phosphide. out of these materials, gaas have highest efficiency, low noise and high operating frequencies. but gaas has a major disadvantage of complex fabrication process and higher cost. so, gaas are not preferred over silicon and germanium. | |
| 105. |
3 IMPATT DIODES, SCHOTTKY BARRIER DIODES, PIN DIODES |
| A. | avalanche multiplication |
| B. | break down of depletion region |
| C. | high reverse saturation current |
| D. | none of the mentioned |
| Answer» A. avalanche multiplication | |
| Explanation: a reverse bias voltage exceeding the breakdown voltage is applied to an impatt diode, a high electric field appears across the n+ p junction. this high field imparts sufficient energy to the holes and also to valence electrons to raise themselves to the conduction band. this results in avalanche multiplication of electron hole pair. | |
| 106. |
To prevent an IMPATT diode from burning, a constant bias source is used to maintain at safe limit. |
| A. | average current |
| B. | average voltage |
| C. | average bias voltage |
| D. | average resistance |
| Answer» A. average current | |
| Explanation: avalanche multiplication is a cumulative process resulting in rapid increase of carrier density. to prevent the diode from burning due to this increased carrier density, a constant bias source is used to maintain average current at safe limit. | |
| 107. |
The number of semiconductor layers in IMPATT diode is: |
| A. | two |
| B. | three |
| C. | four |
| D. | none of the mentioned |
| Answer» C. four | |
| Explanation: impatt diode consists of 4 layers according to the construction. it consists of a p+ region and n+ layers at the two ends. in between these layers, a p type layer and an intrinsic region is sandwiched. | |
| 108. |
The resonant frequency of an IMPATT diode is given by: |
| A. | vd/2l |
| B. | vd/l |
| C. | vd/2πl |
| D. | vdd/4πl |
| Answer» A. vd/2l | |
| Explanation: the resonant frequency of an impatt diode is given by the expression vd/2l. here vd is the carrier drift velocity; l is the length of the intrinsic region in the impatt diode. | |
| 109. |
If the length of the intrinsic region in IMPATT diode is 2 µm and the carrier drift velocity are 107 cm/s, then the drift time of the carrier is: |
| A. | 10-11 seconds |
| B. | 2×10-11 seconds |
| C. | 2.5×10-11 seconds |
| D. | none of the mentioned |
| Answer» B. 2×10-11 seconds | |
| Explanation: the drift time of the carrier is defined as the ratio of length of the intrinsic region to the carrier drift velocity. | |
| 110. |
If the length of the intrinsic region in IMPATT diode is 2 µm and the carrier drift velocity are 107 cm/s, then the nominal frequency of the diode is: |
| A. | 12 ghz |
| B. | 25 ghz |
| C. | 30 ghz |
| D. | 24 ghz |
| Answer» B. 25 ghz | |
| Explanation: nominal frequency is defined as the ratio of the carrier drift velocity to twice the length of the intrinsic region. | |
| 111. |
IMPATT diodes employ impact ionization technique which is a noisy mechanism of generating charge carriers. |
| A. | true |
| B. | false |
| Answer» A. true | |
| Explanation: impatt devices employ impact ionization techniques which is too noisy. hence in order to achieve low noise figure, impact ionization is avoided in baritt diodes. the minority injection is provided by punch through of the intermediate region. | |
| 112. |
An essential requirement for the BARITT diode is that the intermediate drift region be completely filled to cause the punch through to occur. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: an essential requirement for the baritt diode is that the intermediate drift region be completely filled to cause the punch through to the emitter-base junction without causing avalanche breakdown of the base collector junction. | |
| 113. |
If the RMS peak current in an IMPATT diode is 700 mA and if DC input power is 6 watt, with the load resistance being equal to |
| A. | 10.1 % |
| B. | 10.21 % |
| C. | 12 % |
| D. | 15.2 % |
| Answer» B. 10.21 % | |
| Explanation: efficiency of impatt diode is defined as the ratio of output rms power to the input dc power. calculating the rms output power from the given rms current and substituting in the equation of efficiency, the efficiency is 10.21%. | |
| 114. |
If the critical field in a Gunn diode oscillator is 3.2 KV/cm and effective length is 20 microns, then the critical voltage is: |
| A. | 3.2 v |
| B. | 6.4 v |
| C. | 2.4 v |
| D. | 6.5 v |
| Answer» B. 6.4 v | |
| Explanation: critical voltage of a gunn diode oscillator is given by the expression lec where l is the effective length and ec is the critical field. substituting the given values in the above equation, critical voltage is 6.4 volts. | |
| 115. |
The production of power at higher frequencies is much simpler than production of power at low frequencies. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: as frequency increases to the millimeter and sub millimeter ranges, it becomes increasingly more difficult to produce even moderate power with solid state devices, so microwave tubes become more useful at these higher frequencies. | |
| 116. |
Microwave tubes are power sources themselves at higher frequencies and can be used independently without any other devices. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: microwave tubes are not actually sources by themselves, but are high power amplifiers. these tubes are in conjunction with low power sources and this | |
| 117. |
The klystron tube used in a klystron amplifier is a type beam amplifier. |
| A. | linear beam |
| B. | crossed field |
| C. | parallel field |
| D. | none of the mentioned |
| Answer» A. linear beam | |
| Explanation: in klystron amplifier, the electron beam passes through two or more resonant cavities. the first cavity accepts an rf input and modulates the electron beam by bunching it into high density and low density regions. | |
| 118. |
In crossed field tubes, the electron beam traverses the length of the tube and is parallel to the electric field. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: in a crossed field or ‘m’ type tubes, the focusing field is perpendicular to the accelerating electric field. since the focusing field and accelerating fields are perpendicular to each other, they are called crossed field tubes. | |
| 119. |
is a single cavity klystron tube that operates as on oscillator by using a reflector electrode after the cavity. |
| A. | backward wave oscillator |
| B. | reflex klystron |
| C. | travelling wave tube |
| D. | magnetrons |
| Answer» B. reflex klystron | |
| Explanation: reflex klystron is a single cavity klystron tube that operates as on oscillator by using a reflector electrode after the cavity to provide positive feedback via the electron beam. it can be tuned by mechanically adjusting the cavity size. | |
| 120. |
A major disadvantage of klystron amplifier is: |
| A. | low power gain |
| B. | low bandwidth |
| C. | high source power |
| D. | design complexity |
| Answer» B. low bandwidth | |
| Explanation: klystron amplifier offers a very narrow operating bandwidth. this is overcome in travelling wave tube (twt). | |
| 121. |
In a oscillator, the RF wave travels along the helix from the collector towards the electron gun. |
| A. | interaction oscillator |
| B. | backward wave oscillator |
| C. | magnetrons |
| D. | none o the mentioned |
| Answer» B. backward wave oscillator | |
| Explanation: in a backward wave oscillator, the rf wave travels along the helix from the collector towards the electron gun. thus the signal for oscillation is provided by the bunched electron beam itself and oscillation occurs. | |
| 122. |
Magnetrons are microwave devices that offer very high efficiencies of about 80%. |
| A. | true |
| B. | false |
| Answer» A. true | |
| Explanation: magnetrons are capable of very high power outputs, on the order of several kilowatts, and with efficiencies of 80% or more. but disadvantage of magnetron is that they are very noisy and cannot maintain frequency or phase coherence when operated in pulse mode. | |
| 123. |
Klystron amplifiers have high noise output as compared to crossed field amplifiers. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: crossed filed amplifiers have very good efficiencies – up to 80%, but the gain is limited to 10-15 db) in addition, the cfa has a noisier output than either a klystron amplifier or twt. its bandwidth can be up to 40%. | |
| 124. |
is a microwave device in which the frequency of operation is determined by the biasing field strength. |
| A. | vtm |
| B. | gyratron |
| C. | helix bwo |
| D. | none of the mentioned |
| Answer» B. gyratron | |
| Explanation: gyratron is a microwave device in which the frequency of operation is determined by the biasing field strength and the electron velocity, as opposed to the dimensions of the tube itself. this makes the gyrator especially useful for microwave frequencies. | |
| 125. |
2 IMPEDANCE MATCHING |
| A. | swr, first voltage minimum |
| B. | swr, first voltage maximum |
| C. | characteristic impedance, first voltage minimum |
| D. | characteristic impedance, first voltage maximum |
| Answer» A. swr, first voltage minimum | |
| Explanation: with a slotted line, swr and the distance of the first voltage minimum from the load can be measured, from this data, load impedance can be found. | |
| 126. |
A modern device that replaces a slotted line is: |
| A. | digital cro |
| B. | generators |
| C. | network analyzers |
| D. | computers |
| Answer» C. network analyzers | |
| Explanation: although slotted lines used to be the principal way of measuring unknown impedance at microwave frequencies, they have largely been superseded by the modern network analyzer in terms of accuracy, versatility and convenience. | |
| 127. |
If SWR=1.5 with a wavelength of 4 cm and the distance between load and first minima is 1.48cm, then the reflection coefficient is: |
| A. | 0.0126+j0.1996 |
| B. | 0.0128 |
| C. | 0.26+j0.16 |
| D. | none of the mentioned |
| Answer» A. 0.0126+j0.1996 | |
| Explanation: ┌= (swr-1)/ (swr+1). substituting for swr in the above equation for reflection co-efficient, magnitude of the reflection co-efficient is 0.2. to find θ, θ=π+2βlmin, substituting lmin as 1.48cm, θ=86.4⁰. hence converting the polar form of the reflection co-efficient into rectangular co- ordinates, reflection co-efficient is 0.0126+j0.1996. | |
| 128. |
High gain is not achievable at microwave frequencies using BJT amplifiers because: |
| A. | device construction |
| B. | complex architecture |
| C. | ports are not matched at high frequencies |
| D. | none of the mentioned |
| Answer» C. ports are not matched at high frequencies | |
| Explanation: at higher frequencies, if higher bandwidth is desired, a compromise on maximum achievable gain is made. but at these higher frequencies, the ports of the amplifier are not matched to 50 Ω. | |
| 129. |
To flatten the gain response of a transistor: |
| A. | biasing current has to be increased |
| B. | input signal level has to increased |
| C. | increase the operational bandwidth |
| D. | give negative feedback to the amplifier |
| Answer» D. give negative feedback to the amplifier | |
| Explanation: negative feedback can be used to increase the gain response of the transistor, improve the input and output match, and increase the stability of the device. | |
| 130. |
In conventional amplifiers, a flat gain response is achieved at the cost of reduced gain. But this drawback can be overcome by using: |
| A. | balanced amplifiers |
| B. | distributed amplifiers |
| C. | differential amplifiers |
| D. | none of the mentioned |
| Answer» A. balanced amplifiers | |
| Explanation: in conventional amplifiers, a flat gain response is achieved at the cost of reduced gain. but this drawback can be overcome by using balanced amplifiers. this is overcome by using two 900 couplers to cancel input and output reflections from two identical amplifiers. | |
| 131. |
Bandwidth of balanced amplifier can be an octave or more, but is limited by the bandwidth of the coupler. |
| A. | true |
| B. | false |
| Answer» A. true | |
| Explanation: in order to achieve flat gain response, balanced amplifiers use couplers to minimize reflections. but this in turn reduces the bandwidth of the amplifier to the coupler bandwidth. | |
| 132. |
Coupler that is mostly used in balanced amplifiers to achieve the required performance is: |
| A. | branch line coupler |
| B. | wilkinson coupler |
| C. | lange coupler |
| D. | waveguide coupler |
| Answer» C. lange coupler | |
| Explanation: lange couplers are broadband couplers and are compact in size. since the bandwidth of a balanced amplifiers depends on the bandwidth of the coupler used. lange coupler is thus preferred over couplers. | |
| 133. |
Distributed amplifiers offer very high |
| A. | gain |
| B. | bandwidth |
| C. | attenuation |
| D. | none of the mentioned |
| Answer» B. bandwidth | |
| Explanation: distributed amplifiers offer very high bandwidth of about 10 decade. but higher gain cannot be achieved using distributed amplifiers and matching at the ports is very important to achieve higher bandwidth. | |
| 134. |
In distributed amplifiers, all the FET stages in the amplifier are connected in series to one another. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: in distributed amplifiers, cascade of n identical fets have their gates connected to a transmission line having a characteristic impedance of zg with a spacing of lg while the drains are connected to a transmission line of characteristic impedance zd, with a spacing ld. | |
| 135. |
uses balanced input and output, meaning that there are 2 signal lines, with opposite polarity at each port. |
| A. | differential amplifier |
| B. | distributed amplifier |
| C. | balanced amplifier |
| D. | none of the mentioned |
| Answer» A. differential amplifier | |
| Explanation: differential amplifier uses balanced input and outputs, meaning that there are 2 signal lines, with opposite polarity at each port. it has two input ports and one output port. the difference of the 2 input signals is amplified. | |
| 136. |
A major advantage of differential amplifiers is: |
| A. | high gain |
| B. | low input impedance |
| C. | higher output voltage swing |
| D. | none of the mentioned |
| Answer» C. higher output voltage swing | |
| Explanation: differential amplifiers can provide higher voltage swings that are approximately double that obtained with single ended amplifier. | |
| 137. |
Along with a differential amplifier, 1800 hybrid is used both at the input and output. |
| A. | true |
| B. | false |
| Answer» A. true | |
| Explanation: a differential amplifier can be constructed using two single-ended amplifiers | |
| 138. |
is a non linear circuit that converts DC power to an AC waveform of desired frequency based on the oscillator design. |
| A. | attenuator |
| B. | amplifier |
| C. | oscillator |
| D. | none of the mentioned |
| Answer» C. oscillator | |
| Explanation: oscillator is a non linear circuit that converts dc power to an ac waveform. most rc oscillators provide sinusoidal outputs, which minimizes undesired harmonics and noise sidebands. | |
| 139. |
The transfer function of an RF oscillator is given by: |
| A. | a/ (1-ah (ω)) |
| B. | a/ (1+ah (ω)) |
| C. | a/ (-1+ah (ω)) |
| D. | 1/ (1-ah (ω)) |
| Answer» A. a/ (1-ah (ω)) | |
| Explanation: transfer function of an rf oscillator is given by a/ (1-ah (ω)). here, a is the gain of the transistor multiplier used. | |
| 140. |
The criterion on which oscillations are produced in the oscillator circuit is called: |
| A. | shannon’s criteria |
| B. | barkhausen criteria |
| C. | colpitts criteria |
| D. | none of the mentioned |
| Answer» B. barkhausen criteria | |
| Explanation: when the condition 1-ah (ω) | |
| 141. |
The necessary condition for oscillation in a Colpitts oscillator is: |
| A. | c2/c1=gm/gi |
| B. | c1/c2=gm/gi |
| C. | c2/c1= gm*gi |
| D. | none of the mentioned |
| Answer» A. c2/c1=gm/gi | |
| Explanation: the condition for sustained oscillation in a colpitts oscillator is c2/c1 = gm/gi. here c1 and c2 are the capacitance in the feedback network, gm is the transconductance of the transistor and gi is the input admittance. | |
| 142. |
Colpitts oscillator operating at 50 MHz has an inductor in the feedback section of value 0.10µH. then the values of the capacitors in the feedback section is: |
| A. | 100 pf, 100 pf |
| B. | 100 pf, 50 pf |
| C. | 70 pf, 130 pf |
| D. | 80 pf, 60 pf |
| Answer» A. 100 pf, 100 pf | |
| Explanation: the equivalent value of series combination of the capacitors is given by 1/ ω2l. this gives the equivalent capacitance value of 200 pf. c1c2/ (c1+c2) =200 pf. c1 | |
| 143. |
An inductor is operating at frequency of 50 MHz. Its inductance is 0.1 µH, and then the series resistance associated with the inductor is: (Qo=100) |
| A. | 0.31 Ω |
| B. | 1.32 Ω |
| C. | 1 Ω |
| D. | 1.561 Ω |
| Answer» A. 0.31 Ω | |
| Explanation: series resistance associated with an inductor is given by ωl/qₒ. | |
| 144. |
Hartley oscillator has inductance values of 12 mH and 4 mH in the feedback section and a capacitor of 4 nF. Then the resonant frequency of the circuit is: |
| A. | 19.89 khz |
| B. | 25 khz |
| C. | 45 khz |
| D. | 12 khz |
| Answer» A. 19.89 khz | |
| Explanation: resonant frequency of hartley oscillator is given by 1/ 2π√(c1 (l1 + l2)). substituting the given values in the above equation, cut-off frequency is 19.89 khz. | |
| 145. |
Colpitts oscillator in the feedback section has an inductance of 4 mH and capacitors of 12 nH and 4 nH. Then the resonant frequency of Colpitts oscillator is: |
| A. | 50.4 khz |
| B. | 35.1 khz |
| C. | 45.9 khz |
| D. | none of the mentioned |
| Answer» C. 45.9 khz | |
| Explanation: resonant frequency of colpitts oscillator is given by 1/2π√lcₒ, where c0 is the equivalent capacitance given by c1c2/ (c1+c2). substituting and solving the equation, resonant frequency is 45.9 khz. | |
| 146. |
For Colpitts oscillator, the capacitors C1 and C2 in the feedback network are 1 µF and 25 µF respectively. Then the β value of the transistor is: |
| A. | 35 |
| B. | 000.76 |
| C. | 25 |
| D. | 0.0025 |
| Answer» C. 25 | |
| Explanation: β for a transistor is defined as the ratio of transconductance of the transistor to the input admittance, which is equal to the ratio of c2/c1. substituting the given values, β of the transistor is 25. | |
| 147. |
Amplifier efficiency is the ratio of RF output power to DC input power. This parameter determines the performance of an amplifier. |
| A. | true |
| B. | false |
| Answer» A. true | |
| Explanation: power amplifier is the primary consumer of dc power in most hand-held wireless devices, so amplifier efficiency is an important consideration. amplifier efficiency is the ratio of rf output power to dc input power. | |
| 148. |
is defined as the ratio of desired signal power to undesired noise power. |
| A. | signal to noise ratio |
| B. | noise to signal ratio |
| C. | noise figure |
| D. | noise temperature |
| Answer» A. signal to noise ratio | |
| Explanation: snr is defined as the ratio of desired signal power to undesired noise | |
| 149. |
is defined as the ratio of input signal to noise ratio to the output signal to noise ratio. |
| A. | noise figure |
| B. | noise temperature |
| C. | snro |
| D. | none of the mentioned |
| Answer» A. noise figure | |
| Explanation: noise figure is defined as the ratio of input signal to noise ratio to the output signal to noise ratio of a system or a receiver. snri is the signal to noise ratio measured at the input terminals of the device. snr0 is the output signal to noise ratio measured at the output terminals of the device. | |
| 150. |
The equivalent noise temperature of a network given the noise figure of the network or system is: |
| A. | t0(f-1) |
| B. | t0(f+1) |
| C. | t0(f) |
| D. | t0/f |
| Answer» A. t0(f-1) | |
| Explanation: the equivalent noise temperature of a network given the noise figure of the network or system is given by t0(f-1). in this expression, f is the noise figure of the system. t0 has the value 290 k. t0 is the standard temperature considered. | |
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